<h2>Problem 90</h2>
<div style="color:#666;font-size:80%;">04 March 2005</div><br />
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<p>Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.</p>

<p>For example, the square number 64 could be formed:</p>

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<img src="project/images/p_090.gif" alt="" /><br />
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<p>In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49, 64, and 81.</p>

<p>For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.</p>

<p>However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.</p>

<p>In determining a distinct arrangement we are interested in the digits on each cube, not the order.</p>

<p style="margin-left:50px;">{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}<br />
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}</p>

<p>But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers.</p>

<p>How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?</p>
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